The equation of a circle $C$ is $x^2+y^2-14x+4y+28 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Explanation: To find the equation in standard form, complete the square. $(x^2-14x) + (y^2+4y) = -28$ $(x^2-14x+49) + (y^2+4y+4) = -28 + 49 + 4$ $(x-7)^{2} + (y+2)^{2} = 25 = 5^2$ Thus, $(h, k) = (7, -2)$ and $r = 5$.